MATHEMATICS · LESSON 08 OF 10

Rolle and Mean Value Theorems

If a smooth function goes from one value to another, somewhere its instantaneous rate equals its average rate.

What this lesson explains

The Mean Value Theorem (MVT) connects an average rate over an interval with an instantaneous rate at some point inside. It is the reason we can say “if f′ > 0 everywhere, f is increasing”, “if f′ = 0 everywhere, f is constant”, and it gives bounds such as “if the speed never exceeds 90 km/h, then in 2 hours you travel at most 180 km”.

It is also a lesson in reading theorems carefully: the conclusion is guaranteed only when the hypotheses are checked.

Before you begin

Secant slope

The average rate of change of f on [a, b] is f(b) − f(a) / b − a.

Continuity and differentiability

Polynomials, sin, cos and e^x are continuous and differentiable everywhere. Rational functions fail where the denominator is 0; |x| and x^2/3 fail to be differentiable at 0 (corner, cusp).

The idea, made visible

Drive 120 km in 1.5 hours: your average speed is 80 km/h. Common sense says your speedometer must have read exactly 80 km/h at some moment — you cannot average 80 without passing through 80 if your speed changes smoothly. The MVT is the precise version of this.

Geometrically: join the endpoints (a, f(a)) and (b, f(b)) of the graph with a secant line. Somewhere between a and b there is a point c where the tangent line is parallel to the secant, so f′(c) equals the secant slope f(b) − f(a) / b − a.

For f(x) = x²/4 − x/2 + 1 on [0, 4] the secant slope is 0.5. At c = 2 the tangent has the same slope, as the MVT guarantees.

Rolle’s theorem is the special case f(a) = f(b): the secant is horizontal, so somewhere inside the tangent is horizontal, f′(c) = 0. A ball thrown up and caught at the same height has zero vertical velocity at the top.

Both theorems need two hypotheses: f continuous on the closed interval [a, b] and differentiable on the open interval (a, b). Without them the conclusion can fail. |x| on [−1, 1] has f(−1) = f(1), but no horizontal tangent, because of the corner at 0. The theorems promise existence of at least one c; they do not say how many or where. In simple examples we can locate c by solving f′(c) = slope.

Checking hypotheses before using the theorems
Function and interval	continuous on [a, b]?	differentiable on (a, b)?	conclusion guaranteed?
x² on [0, 4]	yes	yes	yes: c = 2
x³ on [0, 2]	yes	yes	yes: c = 2/√3 ≈ 1.155
\|x\| on [−1, 1]	yes	no (corner at 0)	no; indeed no horizontal tangent
1/x on [−1, 1]	no (undefined at 0)	no	no
x^2/3 on [−1, 1]	yes	no (cusp at 0)	no; f′ is never 0

Key terms

Rolle’s theorem: If f is continuous on [a, b], differentiable on (a, b), and f(a) = f(b), then there is at least one c in (a, b) with f′(c) = 0.
Mean Value Theorem: If f is continuous on [a, b] and differentiable on (a, b), then there is at least one c in (a, b) with f′(c) = f(b) − f(a) / b − a.
Hypotheses: The conditions of a theorem (here: continuity on [a, b], differentiability on (a, b), and for Rolle f(a) = f(b)). They must be checked before the conclusion is used.

The formulas and what they mean

Symbols, meanings and units
Symbol	Meaning	Unit
[a, b]	closed interval, endpoints included	input units
(a, b)	open interval, endpoints excluded; c must lie here	input units
c	a point guaranteed by the theorem	input units

Mean Value Theorem

f′(c) = f(b) − f(a) / b − a for some c in (a, b)

Conditions and limits: f continuous on [a, b] and differentiable on (a, b).

Rolle’s theorem

f(a) = f(b) ⇒ f′(c) = 0 for some c in (a, b)

Conditions and limits: Same hypotheses plus equal endpoint values.

Consequences

f′ > 0 on (a, b) ⇒ f increasing; f′ < 0 ⇒ decreasing; f′ = 0 ⇒ f constant

Conditions and limits: On an interval (not across a gap in the domain).

Bounding change

|f(b) − f(a)| ≤ M|b − a| if |f′(x)| ≤ M on (a, b)

Conditions and limits: f satisfies the MVT hypotheses on [a, b].

Why it works: Why f′ > 0 on an interval means f is increasing

Take any two points x₁ < x₂ in the interval.

The MVT on [x₁, x₂] gives c with f(x₂) − f(x₁) = f′(c)(x₂ − x₁).
Rearranged form of the theorem; the hypotheses hold because f is differentiable on the interval.
f′(c) > 0 and x₂ − x₁ > 0, so the right side is positive.
A product of positives is positive.
Therefore f(x₂) > f(x₁).
This is what “increasing” means.

The first-derivative test used for curve sketching rests on this argument.

A first worked example

How to apply Rolle’s theorem or the MVT

Check continuity on the closed interval [a, b]. Look for division by zero, roots of negative numbers and jumps.
Check differentiability on the open interval (a, b). Look for corners (absolute values), cusps and vertical tangents.
For Rolle, check f(a) = f(b). For the MVT, compute the secant slope f(b) − f(a) / b − a.
Solve f′(c) = secant slope (0 for Rolle), and keep only solutions with a < c < b.
If a hypothesis fails, say that the theorem does not apply (the conclusion may or may not still happen).

Find the MVT point for a parabola

Problem. For f(x) = x² on [0, 4], find all c guaranteed by the Mean Value Theorem.

f is a polynomial: continuous on [0, 4] and differentiable on (0, 4).
Hypotheses checked.
Secant slope: 16 − 0 / 4 − 0 = 4.
Average rate of change.
f′(c) = 2c = 4 ⇒ c = 2.
Solve f′(c) = secant slope.
2 lies in (0, 4).
c must be strictly inside the interval.

Result: c = 2.

What it means: For any parabola, the MVT point is the midpoint of the interval.

A different case

An MVT point that is not the midpoint

Problem. For f(x) = x³ on [0, 2], find c.

Polynomial, so the hypotheses hold.
Check first.
Secant slope: 8 − 0 / 2 = 4.
Average rate.
3c² = 4 ⇒ c = ±2 / √3. Only c = 2 / √3 ≈ 1.155 lies in (0, 2).
Reject the negative solution.

Result: c = 2/√3 ≈ 1.155.

What it means: Solving f′(c) = slope may give points outside the interval; they do not count.

More worked cases

Each case below uses a different skill. Every step and result is shown.

Rolle’s theorem

Problem. Verify Rolle’s theorem for f(x) = (x − 2)² on [0, 4].

Polynomial: continuous and differentiable everywhere.
Hypotheses 1 and 2.
f(0) = 4 and f(4) = 4, so f(0) = f(4).
Hypothesis 3.
f′(x) = 2(x − 2) = 0 ⇒ c = 2, which lies in (0, 4).
Solve f′(c) = 0.

Result: c = 2 (the vertex of the parabola).

What it means: The horizontal tangent is at the lowest point, between the two equal endpoint values.

When the theorem does not apply

Problem. f(x) = |x| on [−1, 1] has f(−1) = f(1) = 1. Does Rolle’s theorem guarantee a c with f′(c) = 0?

f is continuous on [−1, 1].
Hypothesis 1 holds.
f is not differentiable at 0, which lies in (−1, 1).
Hypothesis 2 fails: there is a corner.
f′(x) = −1 for x < 0 and +1 for x > 0; it is never 0.
The conclusion is indeed false here.

Result: Rolle’s theorem does not apply, and no such c exists.

What it means: This shows why differentiability is required.

Using the MVT to bound a change

Problem. A car’s speed never exceeds 90 km/h during a 2-hour trip. Use the MVT to bound the distance travelled.

Let s(t) be distance, differentiable, with s′(t) = speed ≤ 90.
Model the trip.
MVT: s(2) − s(0) = s′(c)(2 − 0) for some c in (0, 2).
Apply the theorem on [0, 2].
s′(c) ≤ 90, so s(2) − s(0) ≤ 180.
Bound the instantaneous rate.

Result: At most 180 km.

What it means: Conversely, if the car covered 200 km in 2 h, its speed must have been exactly 100 km/h at some moment.

Common misunderstandings

Misunderstanding: Using the conclusion without checking the hypotheses.
Correct idea: Always state why f is continuous on [a, b] and differentiable on (a, b).
Misunderstanding: Accepting a c outside (a, b).
Correct idea: Solutions of f′(c) = slope outside the open interval are not the ones the theorem guarantees.
Misunderstanding: “The MVT tells us exactly where c is.”
Correct idea: It guarantees existence of at least one c. We can compute c only when f′(c) = slope can be solved.
Misunderstanding: “If the hypotheses fail, there is no such c.”
Correct idea: Then the theorem simply gives no guarantee. The conclusion might still happen by chance.

Keep in mind

f′(c) = f(b) − f(a) / b − a for some c in (a, b)Mean Value Theorem
f(a) = f(b) ⇒ f′(c) = 0 for some c in (a, b)Rolle’s theorem
f′ > 0 on (a, b) ⇒ f increasing; f′ < 0 ⇒ decreasing; f′ = 0 ⇒ f constantConsequences
|f(b) − f(a)| ≤ M|b − a| if |f′(x)| ≤ M on (a, b)Bounding change

Scope of this lesson

Proofs of Rolle’s theorem rely on the Extreme Value Theorem and are sketched only informally here.
The Cauchy (generalised) Mean Value Theorem is not included.

Next: Extrema and curve shape. The first derivative shows where a graph rises and falls; the second shows how it bends. Together they locate maxima, minima and inflection points.

Original study text. Sources and credits.

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